I have the following question:

The heights of $1000$ students are approximately normally distributed with a mean of $174.5$ centimeters and a standard deviation of $6.9$ centimeters. Suppose $200$ random samples of size $25$ are drawn from this population and the means recorded to the nearest tenth of a centimeter. Determine the number of sample means that fall between $172.5$ and $175.8$ centimeters inclusive.

While solving this, the author used $172.45$ and $175.85$, so he changed the values by $0.05$ which I think is continuity correction. However, that is the first time I see continuity correction being used beyond the scope of approximation (my first and only time was with the Binomial-Normal approximation where I understood it easily). I am trying to see why was it used here but it did not make sense to me, here are some parts that I looked in the question to try and figure out if they have a role:

The heights of $1000$ students are approximately normally distributed

I thought that the reason might be this but it did not make much sense to, especially that a lot of previous exercises in another chapter CLT was used without doing this.

the means recorded to the nearest tenth of a centimeter

I tried thinking about this too as since this is rounded to the nearest tenth then adding or removing $0.05$ can potentially make it closer to the real value recorded but this looks like a gamble to me as if for example $172.6$ was rounded to the nearest tenth then adding $0.05$ would make it farther from the real value, so it is a gamble depending on what the real values are which makes it not useful to take.

Determine the number of sample means that fall between $172.5$ and $175.8$ centimeters inclusive

This caught my attention as that is the first time I notice "inclusive" being used by the author in a problem involving the normal distribution only but that doesn't change that $P(X<k)=P(X\leq k)$ in any continuous distribution so I don't think this word should change how we look at the problem.

Any help would be appreciated.


1 Answer 1


Rather than thinking about the question in terms of continuity correction, think about it as a transformed random variable. That is to say, let $X$ be the height random variable for a single student in the population, which we are told is approximately normal with mean $\mu_X = 174.5$ and $\sigma_X = 6.9$. Then the statistic that is being calculated here is $$Y = [\bar X] = \left[ \frac{1}{25} \sum_{i=1}^{25} X_i \right],$$ where $$[ \bar X ] = \frac{1}{10} \begin{cases} \lfloor 10 \bar X \rfloor, & 0 \le 10 \bar X - \lfloor 10 \bar X \rfloor < 0.5 \\ \lceil 10 \bar X \rceil, & 0.5 \le 10 \bar X - \lfloor 10 \bar X \rfloor < 1 \end{cases}$$ denotes the rounded sample mean where the rounding is to the nearest $0.1$. Note that while I have written this function in this elaborate manner, it is not strictly necessary to do this.

The point is that if we understand $Y$ to be a discretizing transformation, then the relevant probability calculation follows naturally: we are interested in the probability that $172.5 \le Y \le 175.8$. But $Y$ has been rounded, so for instance, if $\bar X = 172.45$, then $Y = 172.5$ and this is contained in the desired interval; and if $\bar X = 175.87$, then $Y = 175.9$ which is excluded. So in terms of $\bar X$, we have $$p = \Pr[172.5 \le Y \le 175.8] = \Pr[172.45 \le \bar X < 175.85].$$ Then, the expected number of sample means that fall in this range is simply $200p$, since each sample can be assumed to be nearly independent (in actuality, there is a finite population correction factor involved due to the nonzero probability of resampling the same student(s) more than once across different samples).

The takeaway is that while this could be viewed as a kind of "continuity correction," it isn't really one; rather, it is performing a calculation on a transformed random variable, where the transformation discretizes the underlying continuous variable.

To illustrate with a simpler example, suppose $T \sim \operatorname{Exponential}(1)$ is an exponential random variable with mean $1$; hence its density is $f_T(t) = e^{-t}$, for $t \ge 0$. What happens when we consider the transformed variable $G = \lfloor T \rfloor$? What does $G$ look like? Well, we can immediately see that the support of $G$ is on the nonnegative integers $\{0, 1, 2, \ldots\}$. Moreover, we can easily compute $$\begin{align} \Pr[G = 0] &= \Pr[T < 1] = 1 - e^{-1}, \\ \Pr[G \le 1] &= \Pr[T < 2] = 1 - e^{-2}, \\ \Pr[G \le 2] &= \Pr[T < 3] = 1 - e^{-3}, \\ & \ldots \end{align}$$ so that for each nonnegative integer $n$, $$\Pr[G = n] = \Pr[G \le n] - \Pr[G \le n-1] = (1 - e^{-(n+1)}) - (1 - e^{-n}) = e^{-n} - e^{-(n+1)} = (1 - e^{-1}) e^{-n}.$$ Letting $p = e^{-1}$, we find $$\Pr[G = n] = (1 - p)p^n,$$ which is the probability mass function for a geometric random variable with parameter $p = 1/e$; i.e., $$G \sim \operatorname{Geometric}(1/e).$$ This has nothing to do with continuity correction or even normal (or approximately normal) random variables. Yet if I want to ask $\Pr[2 \le G \le 7]$, we can compute this either via the above PMF we derived, or we can back-transform to the original $T$ and observe $$\Pr[2 \le G \le 7] = \Pr[2 \le T < 8].$$ Such transformations need not result in integer-valued random variables; in your case, $Y$ is not necessarily an integer, but it is nevertheless discrete-valued. Similarly, we can express such a transformation on $T$. For instance, $G' = \frac{1}{100} \lfloor 100 T \rfloor$ takes the greatest hundredth less than or equal to $T$. So if $T = 1.38573$, then $100T = 138.573$, $\lfloor 100T \rfloor = 138$, and $G' = 138/100 = 1.38$. So $G'$ is also discrete and we can compute its probability mass function as well. As an exercise, can you determine whether $G'$ is also geometrically distributed? Why or why not?


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